Template function that can take lambda or function

I maintain an open-source lockless threading library designed for high speed parallel loop unrolling which is used in a couple commercial video games. It has incredibly low-overhead, around 8 clocks for creating a message and around 500 (per thread, including latency) for the entire dispatch and remote execute overhead. I say this first to explain why I don't simply use use std::function and bind.

The library packages a function call and the arguments to the function call in a message (of type Functor<>). Then calls it remotely with copies of the arguments.

I have recently rewritten the library to package remote calls using STL style template meta programming rather than the dated C style macros it originally used. Amazingly this only increased the overhead by two ticks but I cannot figure out how to create a packaging function that takes both lambdas and function pointers.

Desired Usage:

CreateFunctor([](int a, int b){doSomething(a, b)}, 1, 2);

CreateFunctor(&doSomething, 1, 2);

Currently I have to divide these cases into two separate functions, (CreateFunctor and CreateFunctorLambda). If I could combine them I would be able to merge my packaging and dispatching stages into one neat function call and simplify the API.

The problem is that the code for deducing the arguments of a lambda cannot seem to share a template override with the code for deducing the arguments of a function. I've tried using enable_if and it still executes the ::* portion of the lambda version and causes compiler errors with function pointers.

Relevant Snippet:

template <typename... Arguments>
inline Functor<Arguments...> CreateFunctor(void(*func)(Arguments...))
{
    return Functor<Arguments...>(func);
};

template <typename... Arguments>
inline Functor<Arguments...> CreateFunctor(void(*func)(Arguments...), Arguments ... arg)
{
    Functor<Arguments...> ret(func);
    ret.Set(arg...);
    return ret;
};

// template to grab function type that lambda can be cast to
// from http://stackoverflow.com/questions/7943525/is-it-possible-to-figure-out-the-parameter-type-and-return-type-of-a-lambda
template <class T>
struct deduce_lambda_arguments
    : public deduce_lambda_arguments<typename std::enable_if<std::is_class<T>::value, decltype(&T::operator())>::type>
{};

template <class ClassType, typename... Args>
struct deduce_lambda_arguments<void(ClassType::*)(Args...) const>
    // we specialize for pointers to member function
{
    typedef void(*pointer_cast_type)(Args...);
    typedef Functor<Args...> functor_type;
};

template <typename F, typename... Args>
inline auto CreateFunctorLambda(F f, Args... arg) -> typename deduce_lambda_arguments<F>::functor_type
{
    deduce_lambda_arguments<F>::functor_type ret((deduce_lambda_arguments<F>::pointer_cast_type) f);
    ret.Set(arg...);
    return ret;
};

template <typename F>
inline auto CreateFunctorLambda(F f) -> typename deduce_lambda_arguments<F>::functor_type
{
    return deduce_lambda_arguments<F>::functor_type((deduce_lambda_arguments<F>::pointer_cast_type) f);
};

The trick is to ensure that you never evaluate &T::operator() for things-that-you-want-to-support-but-are-not-lambdas. One way to do this is adding an extra template parameter and specializing on that:

template <class T, bool = std::is_class<T>::value>
struct compute_functor_type
    : public compute_functor_type<decltype(&T::operator())>
{};

template <class ClassType, typename... Args>
struct compute_functor_type<void(ClassType::*)(Args...) const, false>
{
    typedef void(*pointer_cast_type)(Args...);
    typedef Functor<Args...> functor_type;
};

template <class ClassType, typename... Args>
struct compute_functor_type<void(ClassType::*)(Args...), false>
{
    typedef void(*pointer_cast_type)(Args...);
    typedef Functor<Args...> functor_type;
};

template <typename... Args>
struct compute_functor_type<void(*)(Args...), false>
{
    typedef void(*pointer_cast_type)(Args...);
    typedef Functor<Args...> functor_type;
};


template <typename F, typename... Args>
inline auto CreateFunctor(F f, Args... arg) 
         -> typename compute_functor_type<F>::functor_type
{
    typename compute_functor_type<F>::functor_type ret((typename compute_functor_type<F>::pointer_cast_type) f);
    ret.Set(arg...);
    return ret;
};

template <typename F>
inline auto CreateFunctor(F f) -> typename compute_functor_type<F>::functor_type
{
    return typename compute_functor_type<F>::functor_type((typename compute_functor_type<F>::pointer_cast_type) f);
};

Saving the (way shorter) original approach here for people who don't have to use MSVC:

Since you only care about captureless non-generic lambdas and convert them to function pointers anyway, this is easy.

Create a matching Functor type from a function pointer:

template<class... Args>
Functor<Args...> make_functor(void (*f)(Args...)) { return {f}; }

And force conversion to function pointer via the unary + operator:

template <class F>
inline auto CreateFunctor(F f) -> decltype(make_functor(+f))
{
    return make_functor(+f);
}

template <class F, typename... Arguments>
inline auto CreateFunctor(F f, Arguments ... arg) -> decltype(make_functor(+f))
{
    auto ret = make_functor(+f);
    ret.Set(arg...);
    return ret;
}